Perhaps you'd be interested in a little more generality.
If $G$ is any finite group (say of order $n$), there are notions of representations of $G$. These are just homomorphisms $\rho:G\to\text{GL}(V)$ where $V$ is a f.d. $\mathbb{C}$-space. Such representations are called irreducible if the only non-trivial $\rho$-invariant (i.e. invariant under all the linear transformations in $\rho(G)\subseteq\text{GL}(V)$) subspaces of $V$ are the trivial ones.
Given any representation $\rho$ of $G$, one can associate to it a character $\chi_\rho:G\to \mathbb{C}$ defined by $\chi_\rho(g)=\text{tr}(\rho(g))$. The set $\text{irr}(G)$ of irreducible characters is merely the set of characters coming from an irreducible representation of $G$.
Now, while it may be non-obvious, the set $\text{irr}(G)$ is finite. In fact, a simplistic bound is that $\#\text{irr}(G)\leqslant n$, and in fact, $\#\text{irr}(G)$ is the number of conjugacy classes of $G$. This comes from the non-obvious fact that if $Z(\mathbb{C}[G])$ denotes the set of class functions on $G$ (i.e. functions $G\to\mathbb{C}$ which are constant on conjugacy classes) then $\text{irr}(G)$ forms a basis for $Z(\mathbb{C}[G])$. In fact, not only do they form a basis, they form an orthonormal basis. Of course, the obvious question, is with respect to what inner product? While it may seem opaque at first, if you haven't seen such matters before, the correct inner product on $Z(\mathbb{C}[G])$ is the weighted convolution:
$$\langle f_1,f_2\rangle=\frac{1}{|G|}\sum_{g\in G}f_1(g)\overline{f_2(g)}$$
Now, there is serious verbiage needed to justify why, in fact, the irreducible characters of $G$ form an orthornormal set with respect to this inner product. But, once this verbiage has been doled out, you get (by mere definition) the following equality
$$\frac{1}{|G|}\sum_{g\in G}\chi_i(g)\overline{\chi_j(g)}=\langle \chi_i,\chi_j\rangle=\delta_{i,j}$$
if $\text{irr}(G)=\{\chi_1,\ldots,\chi_m\}$. This is the so-called first orthogonality relation for irreducible characters.
The first in first orthgonality relation surely hints that we're not done--and we're not. There is a second orthogonality relation:
$$\frac{1}{|G|}\sum_{\chi\in\text{irr}(G)}\chi(g)\overline{\chi(h)}=\#(\mathbf{C}_G(g))c(g,h)$$
where $\mathbf{C}_G(g)$ denotes the centralizer of $g$ in $G$, and $c(g,h)$ is $1$ if $g$ is conjugate to $h$, and zero otherwise.
Now, at this point, you may be really confused as to what this has to do with your question. The answer is somewhat simple, but perhaps non-obvious. Note if $G$ is any group, and $\chi$ is a homomorphism $G\to\mathbb{C}^\times$, then in fact $\chi$ is a representation of $G$--since $\text{GL}_1(\mathbb{C})=\mathbb{C}^\times$. Moreover, since $\mathbb{C}$ has no non-trivial subspaces, such homomorphisms are trivially irreducible! Lastly noting that the trace of something in the range of $G\to\text{GL}_1(\mathbb{C})$ is nothing but the value that $\chi$ takes at the point of $g$, we can see that, in fact, $\text{Hom}(G,\mathbb{C}^\times)\subseteq\text{irr}(G)$. Now, I told you that $\#\text{irr}(G)$ is the number of conjugacy classes of $G$, and so when $G$ is abelian, $\#\text{irr}(G)=n$. But, if $G$ is abelian, say, you know that $\text{Hom}(G,\mathbb{C}^\times)\cong G$ (why?), and thus we can piece everything together to see that $\text{irr}(G)=\text{Hom}(G,\mathbb{C}^\times)$.
Thus, for an abelian group $G$, the orthogonality relations read as follows:
$$\frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\psi(g)}=\delta_{\chi,\psi}\qquad \chi,\psi\in\text{Hom}(G,\mathbb{C}^\times)\qquad\mathbf{(1)}$$
and
$$\frac{1}{|G|}\chi(g)\overline{\chi(h)}=|G|\delta_{g,h}\qquad \chi\in\text{Hom}(G,\mathbb{C}^\times)\qquad\mathbf{(2)}$$
Your two equalities are then special cases of these two identities. Indeed, for your first equality, let $\psi$ be the trivial character, i.e. the trivial map $G\to\mathbb{C}^\times$ in $\mathbf{(1)}$. For your second equality, let $h=1\in G$ in $\mathbf{(2)}$.
You can find proofs for the orthogonality relations in any good book on representation theory.
The above may not be of help to you, being at a possible opaque level of generality, but it's nice to put the theory of Dirichlet characters in the more general context of representation theory of finite groups. This elucidates "why" these identities occur, opposed to cute tricks like the proofs you gave above (which are totally fine, just more opaque). I hope this entices you to read more into the beautiful theory of representation theory. For a simple point of view, taking the attitude similar to what I talked about above, I would highly recommend Steinberg's book on the subject.
Best Answer
The following is a link to a PDF version of a paper published in the Clay Mathematics Proceedings by J. Elstrodt:
https://www.uni-math.gwdg.de/tschinkel/gauss-dirichlet/elstrodt-new.pdf
I didn't see it referenced as a source here or in your posting on Mathematics Overflow, so I am providing it here.
Elstrodt's stated motivation for writing the paper is as follows: "The leading role of German mathematics in the second half of the nineteenth and even up to the faithful year 1933 would have been unthinkable without the foundations laid by Gauss, Jacobi, and Dirichlet. But whereas Gauss and Jacobi have been honoured by detailed biographies (e.g. \ldots), a similar account of Dirichlet's life and work is still a desideritum repeatedly ignored. \ldots The present account is a first attempt to remedy this situation.''
On page 26, the author states, ``According to C.G.J. Jacobi the unit theorem is
one of the most important, but one of the thorniest of the science of number theory.' (references ommitted) adding,Kummer remarks that Dirichlet found the idea of proof when listening to Easter Music in the Sistene Chapel during his Italian journey.''' (and he cites the the 343 reference you alluded to by Kronecker)Now, if one reads the previous two sentences of Elstrodt quickly, he is apt to conclude that the theorem of Dirichlet that the author alludes to is indeed the subject of the first sentence (the Unit Theorem); however, the second sentence does not quite confirm this. And, since the source Elstrodt provides as back-up, as you pointed out in your posting, not not provide convincing evidence as to exactly which theorem of Dirichlet is being referred to.
In any case, I think Elstrodt's 37-page paper may be worth reading as it provides other insights into Kummer's role as an authoritative source of information---this is because (and I am only briefly describing why), Ernst Kummer was married to a cousin of Dirichlet's wife, Rebecca Mendelssohn (sister of composer Felix Mendelssohn), which effectively made Kummer and Dirichlet cousins; whom by the way, had considerable interactions with Dirichlet until the latter's death. It is therefore quite likely that Kummer would have been able to provide many important details in the life of Dirichlet that other biographers would not who were not privy to such information.
Hopefully this in conjunction with the Oct., 2016 edit found on your posting at
https://mathoverflow.net/questions/106196/definitive-source-about-dirichlet-finally-proving-the-unit-theorem-in-the-sistin
adds some support, albeit not the definitive proof you were looking for, that the idea of a proof that Dirichlet had in mind while listening to music in the Sistene Chapel during his Italian journey may very well have pertained to the theorem we now know as Dirichlet's Unit Theorem.